Turbo Machines - Mechanical Engineering (MCQ) questions and answers for Q. 10895

Discussion

• Pankaj   -Posted on 06 Oct 15
  We know that the product of mass and velocity is called as momentum.
  
  ∴ Momentum = m x V
  
  Magnitude of applied force is rate of change of momentum
  
  ∴ Applied force F = d/dt (m x V)
  
  On differentiating the above equation,
  
  F = (m x (dV/dt)) + (V x (dm/dt))
  
  In solid bodies mass of body remains constant,
  
  ∴ dm/dt = 0
  
  F = (m x (dV/dt))
  
  ∴ F = m x dV/dt
  
  = (m/t) (V₂ – V₁ )
  
  Mass flow rate of fluid, ṁ = (m/t) = ρ x Q
  
  Therefore force exerted by the body on fluid is given by,
  
  F = ρ Q (V₂ – V₁ )
  
  Thus the force exerted by fluid on body will be equal and opposite to the force exerted by the body on fluid.
  
  F = ρ Q (V₁ – V₂ )
  
  Therefore,
  
  Force exerted by fluid on body = Rate of momentum in (ṁ V₁) – Rate of momentum out (ṁ V₂)

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